Let $X$ be a smooth variety over some field $k \subset \mathbb{C}$. By a theorem of Grothendieck, one has a canonical isomorphism of complex vector spaces $$ \mathbb{H}^j(X, \Omega_{X/k}^\bullet) \otimes_k \mathbb{C} \stackrel{\sim}{\longrightarrow} \mathbb{H}^j(X(\mathbb{C}), \Omega^\bullet_{X^{an}/\mathbb{C}}) $$ between algebraic and analytic de Rham cohomology.

The situation with coefficients is more subtle. If $\nabla: E \to E \otimes_k \Omega^1_U$ is some integrable connection on a locally free sheaf $E$ on $X$, there is still a canonical morphism $$ \mathbb{H}^j(X, (E \otimes \Omega_{X/k}^\bullet, \nabla)) \otimes_k \mathbb{C} \longrightarrow \mathbb{H}^j(X(\mathbb{C}), (E^{an} \otimes \Omega^\bullet_{X^{an}/\mathbb{C}}, \nabla^{an})) \quad \quad (\ast) $$ between the algebraic (hyper)cohomology of the complex induced by the connection and the corresponding analytic one.

Deligne proves that this is an isomorphism when $\nabla$ has regular singularities.

Question: is this theorem and "if and only if", that is, if the map (*) is an isomorphism, is it true that $\nabla$ has regular singularities?

If this is the case, a reference would be very much appreciated.


1 Answer 1


This should probably be a comment, but it got too long...

Your claim seems to be true if the space $X$ is simply connected. I think it should fail in general, but I am having a hard time coming up with a counterexample.

If $X$ is simply connected, then $(E^{an}, \nabla^{an})$ is isomorphic to the trivial connection. Suppose that

$H^0_{dR}(E,\nabla) \xrightarrow{\sim} H^0_{dR}(E^{an} ,\nabla^{an}) = \mathbb C^r$,

where $r$ is the rank of $E$. Note that $H^0_{dR}(E,\nabla) = Hom_{Conn(X)}((\mathcal O_X,d), (E,\nabla))$, and thus we have a map $f: \mathcal (O_X, d)^{\oplus r} \to (E,\nabla)$ such that the analytification $f^{an}$ is an isomorphism. Hence $f$ is an isomorphism as the analytification functor is conservative. In particular $(E,\nabla) \simeq (\mathcal O_X,d)^{\oplus r}$ is regular.

Thus, in the simply connected case I showed that if analytification induces an isomorphism on $H^0_{dR}$ then the connection is regular.

This stronger statement fails for $X=\mathbb C^\times$. For example, take $E=\mathcal O_X . z^\alpha e^{z}$, where $\alpha \in \mathbb C - \mathbb Z$ (i.e. the trivial line bundle with connection $d + (\alpha/z + 1)dz$). The connection $E$ is non-regular and has no algebraic or analytic flat sections. On the other hand, $H^1_{dR}(E) = \mathbb C$, while $H^1_{dR}(E^{an})=0$ (I think!), so $E$ does not provide a counterexample to your question.

I don't really know much about irregular connections, so I would be curious to see a complete answer to this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.